Racing Brake Pad / Brake system discussion/questions
07-18-2014, 06:30 PM
#241
Drifting
Join Date: Jul 2009
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So... if I understand the OP correctly, all I need to do in order to go faster down the front straight at LS is to drop some weight off the car. Awesome! Who needs horsepower?!?! Now... where is my plasma cutter?
07-21-2014, 12:55 PM
#243
Rennlist Member
Ok, this all doesn't make sense. One last try at explaining it. And it doesn't matter what numbers you use for mass and velocity, the calculations are the same.
if car one has mass of 1 and velocity of 1, then
KE=.5 x mass x Velocity squared or in this case
KE = .5 x 1 x (1x1) = .5
if car two is identical except it has 10% less mass, or .9, what is the velocity ( let's call it z) required to have same KE?
KE =.5 x .9 x (z x z) = .5
If you solve this, z squared = 1/.9
and z = 1.054
so car two would have to increase speed by 5.4% from that 10% weight loss to have the same KE. Any less results in lower KE.
The same calculations can be made for different weight loss, but the end result is that the gain in speed is not likely to be enough to create the same KE.
Therefore, in most situations, a lighter car has less KE, even with the increased velocity that results.
Now to braking,
if car 1 has to brake to velocity .6 to make turn, then after braking, KE = .5 x 1 x (.6x.6) = .18 and KE lost is .5-.18 = .32
if lighter car can carry more speed through turn, there are more variables because we don't know what speed it can carry.
If lighter car carries the 1.054 velocity that gives it the same KE at speed ( which is unlikely), and slows the same 60%, it has a velocity of 1.054x.6=.6324. The KE =.5x.9x(.6324x.6324)=.18 and the KE lost is the same as heavier car at .5-.18=.32 even though car is going faster both at speed and after braking. (Same KE lost, faster lap times)
If we take a more realistic gain in velocity of 3%, then the KE at speed is .5x.9x(1.03x1.03)=.477
If the car slows the same .4 as the heavier car, then we have KE=.5x.9x(.603x.603)= .164 and the KE LOST IS .477-.164=.313 so it actually has to lose less KE to make the turn, despite faster velocity.
If we do it to 60% of the 1.03, then velocity would be 1.03x.6=.618 and KE=.5x.9x(.618x.618)= .172 and the KE lost would be .477-.172=.305 or a lot less.
I know, I know, how can I be sure the lighter car can carry that much more speed into the turn?
Ok, let's do the same calculation and assume the lighter car slows to the same speed as the heavier car, ie .6, then we have
KE=.5x.9x(.6x.6)=.162 and the KE lost is .477-.162=.315. Thus, even when the lighter car slows to the same turn in speed from an initially higher velocity, it has to lose less KE.
That's it, I'm done.
if car one has mass of 1 and velocity of 1, then
KE=.5 x mass x Velocity squared or in this case
KE = .5 x 1 x (1x1) = .5
if car two is identical except it has 10% less mass, or .9, what is the velocity ( let's call it z) required to have same KE?
KE =.5 x .9 x (z x z) = .5
If you solve this, z squared = 1/.9
and z = 1.054
so car two would have to increase speed by 5.4% from that 10% weight loss to have the same KE. Any less results in lower KE.
The same calculations can be made for different weight loss, but the end result is that the gain in speed is not likely to be enough to create the same KE.
Therefore, in most situations, a lighter car has less KE, even with the increased velocity that results.
Now to braking,
if car 1 has to brake to velocity .6 to make turn, then after braking, KE = .5 x 1 x (.6x.6) = .18 and KE lost is .5-.18 = .32
if lighter car can carry more speed through turn, there are more variables because we don't know what speed it can carry.
If lighter car carries the 1.054 velocity that gives it the same KE at speed ( which is unlikely), and slows the same 60%, it has a velocity of 1.054x.6=.6324. The KE =.5x.9x(.6324x.6324)=.18 and the KE lost is the same as heavier car at .5-.18=.32 even though car is going faster both at speed and after braking. (Same KE lost, faster lap times)
If we take a more realistic gain in velocity of 3%, then the KE at speed is .5x.9x(1.03x1.03)=.477
If the car slows the same .4 as the heavier car, then we have KE=.5x.9x(.603x.603)= .164 and the KE LOST IS .477-.164=.313 so it actually has to lose less KE to make the turn, despite faster velocity.
If we do it to 60% of the 1.03, then velocity would be 1.03x.6=.618 and KE=.5x.9x(.618x.618)= .172 and the KE lost would be .477-.172=.305 or a lot less.
I know, I know, how can I be sure the lighter car can carry that much more speed into the turn?
Ok, let's do the same calculation and assume the lighter car slows to the same speed as the heavier car, ie .6, then we have
KE=.5x.9x(.6x.6)=.162 and the KE lost is .477-.162=.315. Thus, even when the lighter car slows to the same turn in speed from an initially higher velocity, it has to lose less KE.
That's it, I'm done.
07-21-2014, 02:26 PM
#244
Drifting
so car two would have to increase speed by 5.4% from that 10% weight loss to have the same KE. Any less results in lower KE.
The same calculations can be made for different weight loss, but the end result is that the gain in speed is not likely to be enough to create the same KE.
07-21-2014, 05:44 PM
#248
Rennlist Member
Thread Starter
wow, i leave out of town for 4 days and you guys are still carriing the torch! thanks! VR is still up to his old antics. posting something he does probably very well.
you're excused. edit: (typo corrected to satisfy VRs high standard of english, but low standards of actual useful knowledge.
)
Did this come out right? the VR is KE. VR is 19.8 Joules. did you mean that the KE was equal and outweighs any effect of the weight if higher speed, or visa versa?
Justin, why do you think it seems "completely unreasonable". go to any drag race, or physics calculator and based on as many factors as you can plug in, if you have 7% less weight 4%increase in speed is certainly in the relm of possiblity. and those of us that have changed our HP/weight by an equal amount or 2x the amount in my case, have seen 10mph as a result.
One thing i forgot to add last week, is dont forget the last turn of laguna, you WOULD be exiting that turn with more speed to start if we are going to play fair with the equations. 4% is probably concervative, and proves the point. (if true, that losing weight is NOT a solution here ) as good as it would be to do it, and has NO downside unless you are trying to solve a overheating braking issue.
I think you are correct, unless i missed something in physics class saying that a better hp/weight ratio doesnt accelerate you at a faster rate.
and that weight will allow you to corner better too!
you're excused. edit: (typo corrected to satisfy VRs high standard of english, but low standards of actual useful knowledge.
)One thing i forgot to add last week, is dont forget the last turn of laguna, you WOULD be exiting that turn with more speed to start if we are going to play fair with the equations. 4% is probably concervative, and proves the point. (if true, that losing weight is NOT a solution here ) as good as it would be to do it, and has NO downside unless you are trying to solve a overheating braking issue.
and that weight will allow you to corner better too!
Last edited by mark kibort; 07-21-2014 at 09:26 PM.
07-21-2014, 05:48 PM
#249
Rennlist Member
Thread Starter
Really Ray, cant you control yourself?
If you are a car guy, maybe even a racer (as i see you have Numbers on your car, so that means your a racer, right). you should be interesting in learning something new, possibly. if not, just go to the wax and wash thread and give them your take on what rags to use.
Try this...... don't say anything.
Go look some of this stuff up, because actually, i bet several people here are actual learning something. I know i am.
If you are a car guy, maybe even a racer (as i see you have Numbers on your car, so that means your a racer, right). you should be interesting in learning something new, possibly. if not, just go to the wax and wash thread and give them your take on what rags to use.
Go look some of this stuff up, because actually, i bet several people here are actual learning something. I know i am.
07-21-2014, 06:02 PM
#250
Rennlist Member
Thread Starter
Ok, this all doesn't make sense. One last try at explaining it. And it doesn't matter what numbers you use for mass and velocity, the calculations are the same.
if car one has mass of 1 and velocity of 1, then
KE=.5 x mass x Velocity squared or in this case
KE = .5 x 1 x (1x1) = .5
if car two is identical except it has 10% less mass, or .9, what is the velocity ( let's call it z) required to have same KE?
KE =.5 x .9 x (z x z) = .5
If you solve this, z squared = 1/.9
and z = 1.054
so car two would have to increase speed by 5.4% from that 10% weight loss to have the same KE. Any less results in lower KE.
The same calculations can be made for different weight loss, but the end result is that the gain in speed is not likely to be enough to create the same KE.
Therefore, in most situations, a lighter car has less KE, even with the increased velocity that results.
Now to braking,
if car 1 has to brake to velocity .6 to make turn, then after braking, KE = .5 x 1 x (.6x.6) = .18 and KE lost is .5-.18 = .32
if lighter car can carry more speed through turn, there are more variables because we don't know what speed it can carry.
If lighter car carries the 1.054 velocity that gives it the same KE at speed ( which is unlikely), and slows the same 60%, it has a velocity of 1.054x.6=.6324. The KE =.5x.9x(.6324x.6324)=.18 and the KE lost is the same as heavier car at .5-.18=.32 even though car is going faster both at speed and after braking. (Same KE lost, faster lap times)
If we take a more realistic gain in velocity of 3%, then the KE at speed is .5x.9x(1.03x1.03)=.477
If the car slows the same .4 as the heavier car, then we have KE=.5x.9x(.603x.603)= .164 and the KE LOST IS .477-.164=.313 so it actually has to lose less KE to make the turn, despite faster velocity.
If we do it to 60% of the 1.03, then velocity would be 1.03x.6=.618 and KE=.5x.9x(.618x.618)= .172 and the KE lost would be .477-.172=.305 or a lot less.
I know, I know, how can I be sure the lighter car can carry that much more speed into the turn?
Ok, let's do the same calculation and assume the lighter car slows to the same speed as the heavier car, ie .6, then we have
KE=.5x.9x(.6x.6)=.162 and the KE lost is .477-.162=.315. Thus, even when the lighter car slows to the same turn in speed from an initially higher velocity, it has to lose less KE.
That's it, I'm done.
if car one has mass of 1 and velocity of 1, then
KE=.5 x mass x Velocity squared or in this case
KE = .5 x 1 x (1x1) = .5
if car two is identical except it has 10% less mass, or .9, what is the velocity ( let's call it z) required to have same KE?
KE =.5 x .9 x (z x z) = .5
If you solve this, z squared = 1/.9
and z = 1.054
so car two would have to increase speed by 5.4% from that 10% weight loss to have the same KE. Any less results in lower KE.
The same calculations can be made for different weight loss, but the end result is that the gain in speed is not likely to be enough to create the same KE.
Therefore, in most situations, a lighter car has less KE, even with the increased velocity that results.
Now to braking,
if car 1 has to brake to velocity .6 to make turn, then after braking, KE = .5 x 1 x (.6x.6) = .18 and KE lost is .5-.18 = .32
if lighter car can carry more speed through turn, there are more variables because we don't know what speed it can carry.
If lighter car carries the 1.054 velocity that gives it the same KE at speed ( which is unlikely), and slows the same 60%, it has a velocity of 1.054x.6=.6324. The KE =.5x.9x(.6324x.6324)=.18 and the KE lost is the same as heavier car at .5-.18=.32 even though car is going faster both at speed and after braking. (Same KE lost, faster lap times)
If we take a more realistic gain in velocity of 3%, then the KE at speed is .5x.9x(1.03x1.03)=.477
If the car slows the same .4 as the heavier car, then we have KE=.5x.9x(.603x.603)= .164 and the KE LOST IS .477-.164=.313 so it actually has to lose less KE to make the turn, despite faster velocity.
If we do it to 60% of the 1.03, then velocity would be 1.03x.6=.618 and KE=.5x.9x(.618x.618)= .172 and the KE lost would be .477-.172=.305 or a lot less.
I know, I know, how can I be sure the lighter car can carry that much more speed into the turn?
Ok, let's do the same calculation and assume the lighter car slows to the same speed as the heavier car, ie .6, then we have
KE=.5x.9x(.6x.6)=.162 and the KE lost is .477-.162=.315. Thus, even when the lighter car slows to the same turn in speed from an initially higher velocity, it has to lose less KE.
That's it, I'm done.
well, the point of all this (and this is really a side bar of the real question as one person said, drop weight to solve the braking issue and a few agreed)
is that if you drop 7% in weight, i think that 4% is reasonable and has been shown to be possible in some of the better performance /simulator calculators.
And if this is true, which it certainly could be, you would have NO gain in braking, and actually it would be slightly worse, in for that system to try and rid its heat. This is because the delta in KE (VRs) is likely going to be the same. ALSO! (and this is a big also) the cornering ability of a lighter car will allow for an increase in exit speed off that final turn down the straight. you might get 2mph just for that alone! added to 3-4mph down the main straight and suddenly you have made things a whole lot worse for the brakes. Great for racing and better lap times, but worse for the brakes ability to shed heat.
07-21-2014, 06:18 PM
#251
Rennlist Member
Thread Starter
a repost of the example i made before.
I think if i added the 2mph exit speed increase off turn 11 for the 200lbs less, that would toss even more "weight" in the corner of the idea, that losing weight doesnt help a braking problem. dream up any numbers you want and i think, worst case, it becomes the same.
anyone without a real understanding of the KE equation, can continue to go on inuition and perception, but in the end, the reality is that speed is greater factor than adding or subtracting weight. weight is linear and speed is squared.
This means the other advice is the way im going. however, those ideas also have some holes too, which i mentioned on this fabilous discussion.
in summary:
I dont bias can carry enough force to slow down that much quicker to avoid the overheat problem.
i also think cooling might not work, as others have agreed, because of the test ive down with cool brakes before this braking point to turn in problem.
im left with trying higher temp pads (Xp20s carbotech) or get bigger rotors and or calipers.
However, just got the high temp paint and brake ducting parts today, so im testing it all out anyway, along with the testing methods suggested here.
Mk
I think if i added the 2mph exit speed increase off turn 11 for the 200lbs less, that would toss even more "weight" in the corner of the idea, that losing weight doesnt help a braking problem. dream up any numbers you want and i think, worst case, it becomes the same.
anyone without a real understanding of the KE equation, can continue to go on inuition and perception, but in the end, the reality is that speed is greater factor than adding or subtracting weight. weight is linear and speed is squared.
This means the other advice is the way im going. however, those ideas also have some holes too, which i mentioned on this fabilous discussion.
in summary:
I dont bias can carry enough force to slow down that much quicker to avoid the overheat problem.
i also think cooling might not work, as others have agreed, because of the test ive down with cool brakes before this braking point to turn in problem.
im left with trying higher temp pads (Xp20s carbotech) or get bigger rotors and or calipers.
However, just got the high temp paint and brake ducting parts today, so im testing it all out anyway, along with the testing methods suggested here.
Mk
This is for VR (Dave), Mike and a few others that have questioned the presentation of the physics behind the conclusion that lightening a car will not help an overheating of the brakes situation.
If you look at the numbers and the logic presented and you find any flaws, please bring them up. I think , most use intuition and experience in making conclusions and decisions in racing. sometimes changes have little or nothing to do with the outcome, because the environments in racing are so dynamic.
Too many examples to list. However, let me frame a picture of what happens at the end of the straight into a braking zone. the entire braking zone and no more.
Conditions: 120mph top speed before brakes applied and speed brought down to 60mph before turn in.
Example: two exact same cars, one with 200lbs removed . (3000lbs and 2800lbs)
Units of measure: Im keeping with "LBS" and "MPH", rather than KG and Meters/sec for mass and velocity. Lets call my new units of measure, "VR"s in the spirit of the discussion. (1) VR is equal to 19.8 J. (Joules).
A Joule is a measure of energy. the numbers are a little bigger, but i think the familarity for the speeds in MPH vs m/s and weight in lbs vs KG is better.
both cars reach a common braking point , call it the #4 marker. they apply brakes to the limit of the tires.
Assumptions: the 7% lighter car reaches a 4% higher speed. (sounds fair right?) so, the 3000lbs car gets to 120mph and the 2800lb car gets to 125mph.
both cars brake down to 60mph from their top speed and turn in.
How much KE was released by the brake. Stoptech says 80% of the energy is released by the braking system to slow the. lets still with 100% for sake of simplcity.
KE=1/2mass x velocity^2
3000lb car at 120mph is 21,600,000 VRs
2800lb car at 125mph is 21,875,000 VRs
(NOTE: if the 2800lb car wanted to sand bag and go the same speed, even though lighter, it would have 7% less KE and that would solve heating issues in most cases)
the 2800lbs car has 1.2% more KE (just as a note here)
Now, if they both slow to 60mph and turn it whats the KE at 60mph for both cars? :
3000lb car at 60mph is 5,400,000
2800lb car at 60mph is 5,040,000
see , again same speed , the difference in 7% mass determines the 7% difference in KE. but hold on.
for both cars to decel to the 60mph, the KE the brakes have to absorb and shed is the difference. Using 2nd grade math, we get:
KE released by decelling from 120 to 60mph 3000lb car:
16,200,000VRs
KE released by decelling from 125 to 60mph 2800lb car:
16,835,000VRs
This proves that the l7% lighter car that brakes to the same speed before turn in, has to deal with 3% more KE. its worse for the brakes! not better by going lighter! AGAIN, if both cars , at their respective weight did this drill from the same speed 120mph to 60, the lighter car would have 7% KE savings. who cares... we know if you go lighter , you will go faster, and faster generates more KE than weight. (see James article). so, this begs the next observation that "well, if the car is lighter, it will not have to slow as much! ". GREAT question, right.....??
So, lets say the 7% lighter car turns in at 4% faster speed. (keeping the effects of acceleration in a straight line, consistent with lateral acceleration. sounds fair right? . So , now the lighter car turns in at 62.5mph.
what happens now?
the lighter car's change in KE to go from 125 to 62.5mph is 16,318,400VRs This is STILL 1% greater than the lighter car.
So, the conclusion:
you lighten the car by 7% its reasonable to expect a 4% greater terminal velocity, and a 4% faster turn in speed. (you can play with the numbers a little and not much will change)
removing weight, in no way , will help your brake cooling or heat release issues you are dealing with the same change in KE with slowing a lighter but faster car vs a heavier slower car. this is usually the way things in physics work. you don't need to dig into thermodynamics Mike, to see the obvious here. Now, are there things you can do to rid heat? sure. ducting, bias, all the things mentioned can change these results to benefit a lighter car for a multitude of problems. The specific problem I'm trying to solve is a heat at the end of a straight, where starting cool or hot doesn't seem to matter. it points to a limit of the components, an no one has offered someone with the same KE issues for a 12.6 rotor. not to mention the smaller caliper size.
this is a sidebar discussion.
if you truly believe that lightening your car willl help with a braking overheating problem, find a flaw in the above presentation. talk to someone that knows physics, racing or just institutional. It seems pretty straight forward to me. But, we all can be wrong or misinterpret the information.
Ive seen it man times before. remember guys that put on a 5 lb lighter rim and see 10hp gain? Just because you see the results, it doesnt mean its due to the mod. so many factors as i said. In the dyno situation, you can actually calibrate the dyno knowing rates of change of acceleration, weight and diameter of wheels. if you know the effects of the wheel lightening is 1hp, and you see 10hp..... its due to something else. you cant fool Mr Watt or newton!
If you look at the numbers and the logic presented and you find any flaws, please bring them up. I think , most use intuition and experience in making conclusions and decisions in racing. sometimes changes have little or nothing to do with the outcome, because the environments in racing are so dynamic.
Too many examples to list. However, let me frame a picture of what happens at the end of the straight into a braking zone. the entire braking zone and no more.
Conditions: 120mph top speed before brakes applied and speed brought down to 60mph before turn in.
Example: two exact same cars, one with 200lbs removed . (3000lbs and 2800lbs)
Units of measure: Im keeping with "LBS" and "MPH", rather than KG and Meters/sec for mass and velocity. Lets call my new units of measure, "VR"s in the spirit of the discussion. (1) VR is equal to 19.8 J. (Joules).
A Joule is a measure of energy. the numbers are a little bigger, but i think the familarity for the speeds in MPH vs m/s and weight in lbs vs KG is better.
both cars reach a common braking point , call it the #4 marker. they apply brakes to the limit of the tires.
Assumptions: the 7% lighter car reaches a 4% higher speed. (sounds fair right?) so, the 3000lbs car gets to 120mph and the 2800lb car gets to 125mph.
both cars brake down to 60mph from their top speed and turn in.
How much KE was released by the brake. Stoptech says 80% of the energy is released by the braking system to slow the. lets still with 100% for sake of simplcity.
KE=1/2mass x velocity^2
3000lb car at 120mph is 21,600,000 VRs
2800lb car at 125mph is 21,875,000 VRs
(NOTE: if the 2800lb car wanted to sand bag and go the same speed, even though lighter, it would have 7% less KE and that would solve heating issues in most cases)
the 2800lbs car has 1.2% more KE (just as a note here)
Now, if they both slow to 60mph and turn it whats the KE at 60mph for both cars? :
3000lb car at 60mph is 5,400,000
2800lb car at 60mph is 5,040,000
see , again same speed , the difference in 7% mass determines the 7% difference in KE. but hold on.
for both cars to decel to the 60mph, the KE the brakes have to absorb and shed is the difference. Using 2nd grade math, we get:
KE released by decelling from 120 to 60mph 3000lb car:
16,200,000VRs
KE released by decelling from 125 to 60mph 2800lb car:
16,835,000VRs
This proves that the l7% lighter car that brakes to the same speed before turn in, has to deal with 3% more KE. its worse for the brakes! not better by going lighter! AGAIN, if both cars , at their respective weight did this drill from the same speed 120mph to 60, the lighter car would have 7% KE savings. who cares... we know if you go lighter , you will go faster, and faster generates more KE than weight. (see James article). so, this begs the next observation that "well, if the car is lighter, it will not have to slow as much! ". GREAT question, right.....??
So, lets say the 7% lighter car turns in at 4% faster speed. (keeping the effects of acceleration in a straight line, consistent with lateral acceleration. sounds fair right? . So , now the lighter car turns in at 62.5mph.
what happens now?
the lighter car's change in KE to go from 125 to 62.5mph is 16,318,400VRs This is STILL 1% greater than the lighter car.
So, the conclusion:
you lighten the car by 7% its reasonable to expect a 4% greater terminal velocity, and a 4% faster turn in speed. (you can play with the numbers a little and not much will change)
removing weight, in no way , will help your brake cooling or heat release issues you are dealing with the same change in KE with slowing a lighter but faster car vs a heavier slower car. this is usually the way things in physics work. you don't need to dig into thermodynamics Mike, to see the obvious here. Now, are there things you can do to rid heat? sure. ducting, bias, all the things mentioned can change these results to benefit a lighter car for a multitude of problems. The specific problem I'm trying to solve is a heat at the end of a straight, where starting cool or hot doesn't seem to matter. it points to a limit of the components, an no one has offered someone with the same KE issues for a 12.6 rotor. not to mention the smaller caliper size.
this is a sidebar discussion.
if you truly believe that lightening your car willl help with a braking overheating problem, find a flaw in the above presentation. talk to someone that knows physics, racing or just institutional. It seems pretty straight forward to me. But, we all can be wrong or misinterpret the information.
Ive seen it man times before. remember guys that put on a 5 lb lighter rim and see 10hp gain? Just because you see the results, it doesnt mean its due to the mod. so many factors as i said. In the dyno situation, you can actually calibrate the dyno knowing rates of change of acceleration, weight and diameter of wheels. if you know the effects of the wheel lightening is 1hp, and you see 10hp..... its due to something else. you cant fool Mr Watt or newton!
07-21-2014, 06:46 PM
#252
Rennlist Member
You are way off base thinking a 7% weight loss will give you 4% increase in speed. You are saying that my 968 at Watkins Glen at the end of the back straight should go from a weight of 2800 lbs (roughly) and speed of 135 to over 140 if I can get the weight down to 2600. Neither one of those two things will ever happen. Just not realistic.
07-21-2014, 07:03 PM
#253
Rennlist Member
Mark,
The word you are looking for is "you're". When you make fundamental mistakes of just plain basic grade school ignorance like this again and again and agin in the same thread, your [sic] claims of superior intellect/understanding of physics/driving/etc sort of become a laughable puff of smoke.
In addition, Mike Shust destroyed your [sic] argument from a physics and science perspective.
So...where does that leave you?
The word you are looking for is "you're". When you make fundamental mistakes of just plain basic grade school ignorance like this again and again and agin in the same thread, your [sic] claims of superior intellect/understanding of physics/driving/etc sort of become a laughable puff of smoke.
In addition, Mike Shust destroyed your [sic] argument from a physics and science perspective.
So...where does that leave you?
07-21-2014, 08:46 PM
#255
Rennlist Member
Thread Starter
You are way off base thinking a 7% weight loss will give you 4% increase in speed. You are saying that my 968 at Watkins Glen at the end of the back straight should go from a weight of 2800 lbs (roughly) and speed of 135 to over 140 if I can get the weight down to 2600. Neither one of those two things will ever happen. Just not realistic.
I'm also talking HP to weight here. if your 968, isn't pushing 8:1 HP/weight, and we have aero to factor in. you're in left field and you are absolutely right. it wont do it . (your 968)
I've already done the empirical testing. I have some pretty clear video of a car that hasn't changed in 8 years, all of the sudden, the only change is HP. 50hp added, and you can see the 3rd gear shift go from 118 , about 2-3mph from a 4th gear shift, well past the start finish, to 118mph (redline 3rd, so there is no guessing) at the start finish, exactly. that's 50hp. just do the HP to weight calculations and you see that what i put down at a "guess" is a pretty educated one. PLUS, plug it in a simulator. Your'e going to find something in the same ball park. Or, do the calculations on the KE (rate of change) raw.